Solve 3x^2-8x+4

$$3 { x }^{ 2 } -8x+4$$

Factor

$$\left(x-2\right)\left(3x-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-8 ab=3\times 4=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-6 b=-2$$

Rewrite 3x^{2}-8x+4 as \left(3x^{2}-6x\right)+\left(-2x+4\right).

$$\left(3x^{2}-6x\right)+\left(-2x+4\right)$$

Factor out 3x in the first and -2 in the second group.

$$3x\left(x-2\right)-2\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(3x-2\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$3x^{2}-8x+4=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3\times 4}}{2\times 3}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 3\times 4}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-8\right)±\sqrt{64-12\times 4}}{2\times 3}$$

Multiply -12 times 4.

$$x=\frac{-\left(-8\right)±\sqrt{64-48}}{2\times 3}$$

Add 64 to -48.

$$x=\frac{-\left(-8\right)±\sqrt{16}}{2\times 3}$$

Take the square root of 16.

$$x=\frac{-\left(-8\right)±4}{2\times 3}$$

The opposite of -8 is 8.

$$x=\frac{8±4}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{8±4}{6}$$

Now solve the equation x=\frac{8±4}{6} when ± is plus. Add 8 to 4.

$$x=\frac{12}{6}$$

Divide 12 by 6.

$$x=2$$

Now solve the equation x=\frac{8±4}{6} when ± is minus. Subtract 4 from 8.

$$x=\frac{4}{6}$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{2}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and \frac{2}{3} for x_{2}.

$$3x^{2}-8x+4=3\left(x-2\right)\left(x-\frac{2}{3}\right)$$

Subtract \frac{2}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}-8x+4=3\left(x-2\right)\times \frac{3x-2}{3}$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3x^{2}-8x+4=\left(x-2\right)\left(3x-2\right)$$

Evaluate

$$\left(x-2\right)\left(3x-2\right)$$